Given the reaction :
`C(s)+H_2O(l)toCO(g)+H_2(g)`
Calculate the volume of the gases prodcued at STP from 48.0 g of carbon.

Options

A. 181.6 L
B. 179.2 L
C. 45.4 L
D. 22.7 L

Answer

To solve the problem of calculating the volume of gases produced at STP from 48.0 g of carbon in the reaction:

\[ C(s) + H_2O(l) \rightarrow CO(g) + H_2(g) \]

we will follow these steps:

### Step 1: Calculate the moles of carbon
To find the number of moles of carbon, we use the formula:

\[ \text{Moles of carbon} = \frac{\text{mass of carbon}}{\text{molar mass of carbon}} \]

Given:
- Mass of carbon = 48.0 g
- Molar mass of carbon (C) = 12 g/mol

Calculating the moles:

\[ \text{Moles of carbon} = \frac{48.0 \, \text{g}}{12 \, \text{g/mol}} = 4 \, \text{moles} \]

### Step 2: Determine the moles of gases produced
From the balanced chemical equation, we can see that:
- 1 mole of carbon produces 1 mole of CO and 1 mole of H2.

Thus, for 4 moles of carbon:
- Moles of CO produced = 4 moles
- Moles of H2 produced = 4 moles

Total moles of gas produced:

\[ \text{Total moles of gas} = \text{moles of CO} + \text{moles of H2} = 4 + 4 = 8 \, \text{moles} \]

### Step 3: Calculate the volume of gases at STP
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the volume of the gases produced can be calculated as follows:

\[ \text{Volume of gases} = \text{Total moles of gas} \times \text{Volume per mole at STP} \]

Calculating the volume:

\[ \text{Volume of gases} = 8 \, \text{moles} \times 22.4 \, \text{L/mol} = 179.2 \, \text{liters} \]

### Final Answer
The volume of gases produced at STP from 48.0 g of carbon is **179.2 liters**.

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